Integration Problem (Substitution)

**jessicadowning** · October 14th 2012, 09:25 PM

Here's the problem: indefinite integral of (-3x^2)/sqrt((t^3)-16)

**MarkFL** · October 14th 2012, 09:35 PM

Are we integrating with respect to x or t? I suspect you mean for the two variables to be the same, and if so, let u be the value under the radical, and you will find you can easily get du as part of the integral.

**jessicadowning** · October 14th 2012, 10:00 PM

Yes, I meant them to be the same, oops!

I got the answer to be sqrt((t^3)+16)/2 + C, but I don't think that's correct...

**MarkFL** · October 14th 2012, 10:14 PM

If you are unsure of your result, use differentiation to check your result:

$\frac{d}{dx}\left(\frac{\sqrt{t^3+16}}{2}+C \right)=\frac{3t^2}{4\sqrt{t^3+16}}\ne\frac{-3t^2}{\sqrt{t^3+16}}$

We can see it is close (essentially the wrong constant factor of the radical), but not quite right. I would let:

$u=t^3+16\,\therefore\,du=3t^2\,dt$ and so we have:

$-\int u^{-\frac{1}{2}}\,du=-2u^{\frac{1}{2}}+C=-2\sqrt{t^3+16}+C$

Now, checking by differentiation, we find:

$\frac{d}{dx}\left(-2\sqrt{t^3+16}+C \right)=\frac{-3t^2}{2\sqrt{t^3+16}}$

The derivative of the anti-derivative is the original integrand, so our result is correct.

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